Multiply the following rational expressions and simplify the result. $\dfrac{z^2+9y^2z}{5z} \cdot \dfrac{z^2-9yz+18y^2}{z^2-9y^2}=$
Answer: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $z^2+9y^2z$, of the first expression can be factored as $z(z+9y^2)$ by factoring out a $z$. The denominator, $5z$, of the first expression cannot be factored further. The numerator, $z^2-9yz+18y^2$, of the second expression can be factored as $(z-6y)(z-3y)$ using the sum-product pattern. The denominator, $z^2-9y^2$, of the second expression can be factored as $(z+3y)(z-3y)$ using the difference of squares pattern. Now the product looks as follows: $\dfrac{z(z+9y^2)}{5z} \cdot \dfrac{(z-6y)(z-3y)}{(z+3y)(z-3y)}$ To multiply two rational expressions, we multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{z(z+9y^2)}{5z} \cdot \dfrac{(z-6y)(z-3y)}{(z+3y)(z-3y)} \\\\\\ &= \dfrac{z(z+9y^2) \cdot (z-6y)(z-3y)}{5z \cdot (z+3y)(z-3y)} &\text{Multiply across.}\\\\\\\\ &= \dfrac{{\cancel{z}}(z+9y^2) (z-6y){\cancel{(z-3y)}}}{5{\cancel{z}} (z+3y){\cancel{(z-3y)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{(z+9y^2) (z-6y)}{5(z+3y)} \end{aligned}$ Therefore, the simplified form of the product is $\dfrac{(z+9y^2) (z-6y)}{5(z+3y)}$, which is equivalent to $\dfrac{z^2-6yz+9y^2z-54y^3}{5z+15y}$.